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(C) Given the differential equation: $e^{y} \frac{d y}{d x}-2 e^{y} \sin x=-\sin x \cos ^{2} x$. Let $e^{y}=t$,then $e^{y} \frac{d y}{d x}=\frac{d t}{

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$4$

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(C) Given the differential equation: $e^{y} \frac{d y}{d x}-2 e^{y} \sin x=-\sin x \cos ^{2} x$. Let $e^{y}=t$,then $e^{y} \frac{d y}{d x}=\frac{d t}{d x}$. The equation becomes $\frac{d t}{d x}-2 \sin x \cdot t=-\sin x \cos ^{2} x$. This is a linear differential equation of the form $\frac{d t}{d x}+P(x)t=Q(x)$,where $P(x)=-2 \sin x$ and $Q(x)=-\sin x \cos ^{2} x$. The integrating factor $I.F. = e^{\int -2 \sin x dx} = e^{2 \cos x}$. The solution is $t \cdot e^{2 \cos x} = \int -\sin x \cos ^{2} x \cdot e^{2 \cos x} dx + C$. Let $u = \cos x$,then $du = -\sin x dx$. The integral becomes $\int u^{2} e^{2u} du$. Using integration by parts: $\int u^{2} e^{2u} du = u^{2} \frac{e^{2u}}{2} - \int 2u \frac{e^{2u}}{2} du = \frac{u^{2} e^{2u}}{2} - (u \frac{e^{2u}}{2} - \int \frac{e^{2u}}{2} du) = \frac{u^{2} e^{2u}}{2} - \frac{u e^{2u}}{2} + \frac{e^{2u}}{4}$. So,$e^{y} e^{2 \cos x} = e^{2 \cos x} (\frac{\cos^{2} x}{2} - \frac{\cos x}{2} + \frac{1}{4}) + C$. At $x = \frac{\pi}{2}$,$y = 0$,so $e^{0} e^{0} = e^{0} (0 - 0 + \frac{1}{4}) + C \Rightarrow 1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$. Thus,$e^{y} = \frac{\cos^{2} x}{2} - \frac{\cos x}{2} + \frac{1}{4} + \frac{3}{4} e^{-2 \cos x}$. At $x = 0$,$e^{y(0)} = \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{3}{4} e^{-2} = \frac{1}{4} + \frac{3}{4} e^{-2}$. Comparing with $\alpha + \beta e^{-2}$,we get $\alpha = \frac{1}{4}$ and $\beta = \frac{3}{4}$. Therefore,$4(\alpha + \beta) = 4(\frac{1}{4} + \frac{3}{4}) = 4(1) = 4$.

Let $y=y(x)$ be the solution of the differential equation $e^{y} \frac{d y}{d x}-2 e^{y} \sin x+\sin x \cos ^{2} x=0$,with $y(\frac{\pi}{2})=0$. If $y(0)=\log _{e}(\alpha+\beta e^{-2})$,then $4(\alpha+\beta)$ is equal to $....$

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